Lab 12: Generative recursion 2

Remember to follow (and when necessary, adapt) the design recipe for each function. When you write a generative recursive function, you must provide a termination argument.

1 Binary search

Searching is a very common task: we might want to know how many times a given word occurs in a book, or find someone’s phone number in a contact list or phonebook. In Problem set 7: Lists, you implemented searching through a list of frequencies. In Problem set 10: Prefix trees, you implemented searching through a tree of letters. A list and a tree represent two different ways to decompose a search problem. Searching through a list can be slower than searching through a tree, because in a list, the target can be buried much farther from the root.

Binary search is a general and efficient search algorithm. It requires the data to be sorted, like the headwords in a dictionary. To look for a word in a dictionary using binary search, first we flip to the middle of the dictionary, and compare the word in the middle to the word we want. If they happen to be the same, then we’re lucky and done. If they’re different, then depending on their relative order, we can always throw away half of the dictionary. For instance, if we’re looking for “bro” and the middle of the dictionary is “law”, then we can throw away the second half of the dictionary, and continue with the middle of the first half.

Exercise 3. Design a function search-tree-contains? that determines whether the given SearchTree contains the given string. For example, neighbors contains "Monroe" but not "Middlesex".

Instead of generative recursion, stick with structural recursion and follow the template for processing a SearchTree. Do not search through the entire tree; that’s way slower. Instead, assume that the tree is sorted. So, every time you come to a try, you can compare its middle against the given string (using string=? and string<?) to decide whether to go left, stop, or go right. The video above only makes 3 comparisons! In other words, never make both of the two recursive calls offered by the template.

The goal of the rest of this section is to generate a SearchTree from a sorted list of county names. The problem decomposition can use some helper functions.

Exercise 6. Design a function right-half that takes a non-empty list as input and returns its second half without its middle element. If the input list has 9 elements, then the output list should be its last 4 elements. If the input list has 4 elements, then the output list should contain just its last element.

Exercise 7. Design a function generate-search-tree that decomposes a given [ListOf String] into a SearchTree. Follow the decomposition strategy in Exercise 1.

Hint: Use generative recursion. Again, remember to make your function terminate, and explain how by writing a termination argument. Use left-half, middle-element, and right-half.

Challenge. There’s a bug! Decatur is an Indiana county name, and it is in the text file, but search-tree-contains? fails to find it. What’s causing this bug? How can you fix it without reordering the names?

2 Euclid’s algorithm

The greatest common divisor (GCD) of two numbers is the largest number that divides them both. For example, the GCD of 115 and 50 is 5, because 5 divides both 115 and 50 evenly, and no number greater than 5 divides both 115 and 50 evenly. In this section, you’ll use generative recursion to find the GCD of two numbers, without using any built-in function for doing so. You will implement Euclid’s algorithm, which is more than 2000 years old!

Exercise 10. Now that we have decomposed the GCD problem, we can read off the answer by processing the decomposition.

Exercise 11. To represent Euclid’s algorithm by a picture, design a function euclid-tree-rectangle that takes a EuclidTree and returns an image whose width is the numerator and whose height is the denominator. The image should look like a rectangle made of a nest of squares. For example, the picture to the right represents computing the GCD of 115 and 50. It is 115 pixels wide and 50 pixels tall. It contains 2 large squares, of size 50, because dividing 115 by 50 gives the quotient 2. It also contains 3 medium squares, of size 15, because dividing 50 by 15 gives the quotient 3. Finally it contains 3 small squares, of size 5, because dividing 15 by 5 gives the quotient 3.